Most COMMON misconception: Topic of another video but the gravitational mass is unaffected by linear velocity, which is a VERY common misconception. Only how hard it is to "push" an object, i.e. it's inertia. Otherwise you get crazy predictions like accelerating a proton to close to the speed of light and it exerting near infinite gravitational force! : )
For the coders out there btw: I coded the animations using Manim in Python, the software created by 3blue1brown. I am blown away by its potential for visualisations in physics!
I am always amazed how little calculus there is in SR! Probably a testament to the fact we are dealing with intertial frames of reference. Thanks for the comment!
There is an 'i' in momentum. E=mc^2 is hyperbolic. cos(arctan(i*y/x))=c/sqrt(x^2+y^2) γ(x,y)-> x/sqrt(x^2+y^2) γ(i*v, c)=1/sqrt(1-(v/c)^2) sin(arctan(i*v/c))=i*v/sqrt(c^2-v^2) p=mvγ/c=m*v*cos(arctan(i*v/c))=m*c*v/sqrt(c^2-v^2) the v in p has the 'i' i*v/sqrt(c^2-v^2)≠m*c*v/sqrt(c^2-v^2) There is also a limit to γ, it is sqrt(c^2-2). Anything greater than v=c-1/c is meaningless. cos(arctan(i*sqrt(c^2-2)/c))+sin(arctan(i*sqrt(c^2-2)/c))=c*sqrt(2)/2+i*c*sqrt(2)/2. This is 45 degrees. The function continues at -45 degrees, so anything greater than sqrt(c^2-2) and less than c is meaningless. @-45 degrees and moves to zero, the value @-45 degrees is c*sqrt(2)/2-i*c*sqrt(2)/2 (c*sqrt(2)/2+i(c*sqrt(2)/2)+c*sqrt(2)/2-i(c*sqrt(2)/2))/2=c*sqrt(2)/2 (c*sqrt(2)/2+i(c*sqrt(2)/2))(c*sqrt(2)/2-i(c*sqrt(2)/2))=c^2
For someone who has the standard model lagrangian on his shirt, you should've known before making this video that mass is Lorentz invariant and as such there's no thing like relativistic mass. Mass is the same in every inertial frame and therefore a body at rest with mass m has the same mass m when moving near to the light speed.
Completely agree, btw, I took extra care to call it inertia. In the case of the binomial expansion it is pretty useful to give the first term a name and the rest of the energy expansion a name. Thanks for the feedback!!
It comes from momentum p = m*v. If two bodies with the same mass collide with each other elastically. The overall momentum will be Zero. A) Collision ---------->00 use the binomial expansion used in this vid. and you get E = mc^2 So squaring c was already there during the deriving of the Lorenz factor 1/(1-v^2/c^2). In this case with the example I used with Pythagoras. But this is a simple example.
@@zhelyo_physicsThank you! I noticed something, in m= m0/sqrt(1-v^2/c^2 m= m0×(1-v^2/c^2)^½ is wrong because its m0×(1-v^2/c^2)^-½ which I calculated (just for safe) and its actually in ^-½
Not into geometry? Here's a very basic question that should give you pause. Why do you claim -6×-6 = 36 ? Or that 1/0 = i ? Because you're too proud to admit you just don't know and never will
BTW those crazy reaches of logic were all made when the numbers weren't fitting the narrative just use letters. If they ask you how fast is electricity you just say a big number and we'll find a way to make it fit
Wonderful explanation,but idk if i am right but mc²=m0c²+m0v²/2 must be super close to E but not exactly,to me it seems that it would be true in partial relativity but in total relativity,that thing is super close but not exact,but wonderful explanation man,keep it up,such a complex concept explained so simply yet mathematicaly.😮😮
Very good question. You are correct..for small speeds, v. The idea is the first term does not depend on v, and each subsequent term will get bigger and bigger as v increases. Hope this helps!
Oh ok now i know it thanks but the length contraction was one of main assumptions in relativity it was assumed to be parallel in direction of motion to be able to get gamma if we don't assume that we won't get the formula for time diluation also @zhelyo_physics
interesting point. Actually not. Time dilation and hence the gamma factor is a consequence of assuming Einstein's Postulates of Special Relativity. Once you have the gamma factor you can derive Length Contraction, The Lorentz Transformations etc. I have all the derivations here if you are interested: th-cam.com/video/nUymnmiIZo4/w-d-xo.htmlsi=jM6BevXfDeTlrUaz. Have a great day!
@@zhelyo_physics I don't mean that the formula was one of the main postulates I mean the assumption that length only contract in parallel direction of motion if he didn't assume this we won't use the train and light beam experiment to get gamma as the vertical length of train in its rest frame won't equal the one in the observer frame so we need to assume that first
Not need to be hard. Rubbish. You use the fact that m is related to m_0 via the Lorentz factor (go ahead and derive this) and then use the binominal expansion (again not easy for people not fluent in mathematics). I understood this derivation (and guessed it before the video started) but that's because I have a MSci in Mathematical Physics.
no problem, I appreciate the perspective a lot. My general audience is physics students who are familiar with the above. I am still amazed that most of relativity is based on geometry, algebra and some postulates which I find remarkable. Comparable to other theories that have changed our views of the universe, the mathematical simplicity of this one I find extremely elegant. Thank you for the comment.
I want you solve one question of relativity I was unable to solve this problem Book: Robert Resnick Relativity Page: 48 , Problem:14 , Binary system Please sir my humble request with you , no one was help me 😥🥺
Most COMMON misconception: Topic of another video but the gravitational mass is unaffected by linear velocity, which is a VERY common misconception. Only how hard it is to "push" an object, i.e. it's inertia. Otherwise you get crazy predictions like accelerating a proton to close to the speed of light and it exerting near infinite gravitational force! : )
For the coders out there btw: I coded the animations using Manim in Python, the software created by 3blue1brown. I am blown away by its potential for visualisations in physics!
wow no calculus is needed!
I love this derivation!
I am always amazed how little calculus there is in SR! Probably a testament to the fact we are dealing with intertial frames of reference. Thanks for the comment!
@@zhelyo_physics absolutely!
There is an 'i' in momentum. E=mc^2 is hyperbolic.
cos(arctan(i*y/x))=c/sqrt(x^2+y^2)
γ(x,y)-> x/sqrt(x^2+y^2)
γ(i*v, c)=1/sqrt(1-(v/c)^2)
sin(arctan(i*v/c))=i*v/sqrt(c^2-v^2)
p=mvγ/c=m*v*cos(arctan(i*v/c))=m*c*v/sqrt(c^2-v^2)
the v in p has the 'i'
i*v/sqrt(c^2-v^2)≠m*c*v/sqrt(c^2-v^2)
There is also a limit to γ, it is sqrt(c^2-2). Anything greater than v=c-1/c is meaningless.
cos(arctan(i*sqrt(c^2-2)/c))+sin(arctan(i*sqrt(c^2-2)/c))=c*sqrt(2)/2+i*c*sqrt(2)/2.
This is 45 degrees. The function continues at -45 degrees, so anything greater than sqrt(c^2-2) and less than c is meaningless.
@-45 degrees and moves to zero, the value @-45 degrees is c*sqrt(2)/2-i*c*sqrt(2)/2
(c*sqrt(2)/2+i(c*sqrt(2)/2)+c*sqrt(2)/2-i(c*sqrt(2)/2))/2=c*sqrt(2)/2
(c*sqrt(2)/2+i(c*sqrt(2)/2))(c*sqrt(2)/2-i(c*sqrt(2)/2))=c^2
Could you explain why the higher order terms are negligible and omitted?
Hi ZPhysics, I love differential equations in physics, can you make more videos about them?
For sure! Me too! 😀 have lots of things planned, stay tuned!
For someone who has the standard model lagrangian on his shirt, you should've known before making this video that mass is Lorentz invariant and as such there's no thing like relativistic mass. Mass is the same in every inertial frame and therefore a body at rest with mass m has the same mass m when moving near to the light speed.
Completely agree, btw, I took extra care to call it inertia. In the case of the binomial expansion it is pretty useful to give the first term a name and the rest of the energy expansion a name. Thanks for the feedback!!
E= mc^2
E=? M=? C =?
What's the answer?😊
Why speed of light square?
seems to come naturally out of the expansion, very interesting.
Momentum in the ether medium is an mv^2 function.
It was actually derived, prior to Einstein plagiarizing it, from fluid dynamics.
It comes from momentum p = m*v.
If two bodies with the same mass collide with each other elastically. The overall momentum will be Zero.
A) Collision
---------->00 use the binomial expansion used in this vid. and you get E = mc^2
So squaring c was already there during the deriving of the Lorenz factor 1/(1-v^2/c^2). In this case with the example I used with Pythagoras. But this is a simple example.
Is there a easy or simple definition for inertia because it sometimes confuses me when out into other contexts
yep! F=ma, m=inertial mass.
This is your Newtonian mass and in relativity it is essentially your rest mass, m_0 in the video. Hope this helps!
Thanks I get it now 🙏
Mc2=e ?
There is no such thing as relativistic mass and rest mass, just mass.
I completely agree
What's the difference between ,,m0" and ,,m"
m0 is the rest mass and m is a term which changes the momentum as the speed changes. I call it inertia in the video I think.
@@zhelyo_physicsThank you!
I noticed something, in m= m0/sqrt(1-v^2/c^2
m= m0×(1-v^2/c^2)^½ is wrong because its m0×(1-v^2/c^2)^-½ which I calculated (just for safe) and its actually in ^-½
Poor explanation accompanied by a ridiculously shouty delivery 😢
thank you for the feedback. It is genuinely useful to hear.
Thank you and 'subbed'. Great content!
thank you very much for the comment!
Not into geometry? Here's a very basic question that should give you pause. Why do you claim -6×-6 = 36 ? Or that 1/0 = i ? Because you're too proud to admit you just don't know and never will
Lol you fool it's the root of -1 that is i!
And E=mc^2 just prove it prove it in a lab
Its all bs
BTW those crazy reaches of logic were all made when the numbers weren't fitting the narrative just use letters. If they ask you how fast is electricity you just say a big number and we'll find a way to make it fit
What are you going on about???
Wonderful explanation,but idk if i am right but mc²=m0c²+m0v²/2 must be super close to E but not exactly,to me it seems that it would be true in partial relativity but in total relativity,that thing is super close but not exact,but wonderful explanation man,keep it up,such a complex concept explained so simply yet mathematicaly.😮😮
Very good question. You are correct..for small speeds, v. The idea is the first term does not depend on v, and each subsequent term will get bigger and bigger as v increases. Hope this helps!
It's not pronounced Einstein. It's like Einshtein, why people keep saying this?
How could man know the speed of light? Start there now tell me how he proved that this was any more than a pipe dream
Yo proved something with something came from it
Not really, the concept of the gamma factor is introduced first and then gradually length contraction and ideas of momentum. Hope this helps!
Oh ok now i know it thanks but the length contraction was one of main assumptions in relativity it was assumed to be parallel in direction of motion to be able to get gamma if we don't assume that we won't get the formula for time diluation also @zhelyo_physics
interesting point. Actually not. Time dilation and hence the gamma factor is a consequence of assuming Einstein's Postulates of Special Relativity. Once you have the gamma factor you can derive Length Contraction, The Lorentz Transformations etc.
I have all the derivations here if you are interested: th-cam.com/video/nUymnmiIZo4/w-d-xo.htmlsi=jM6BevXfDeTlrUaz. Have a great day!
@@zhelyo_physics I don't mean that the formula was one of the main postulates I mean the assumption that length only contract in parallel direction of motion if he didn't assume this we won't use the train and light beam experiment to get gamma as the vertical length of train in its rest frame won't equal the one in the observer frame so we need to assume that first
Not need to be hard. Rubbish. You use the fact that m is related to m_0 via the Lorentz factor (go ahead and derive this) and then use the binominal expansion (again not easy for people not fluent in mathematics). I understood this derivation (and guessed it before the video started) but that's because I have a MSci in Mathematical Physics.
no problem, I appreciate the perspective a lot. My general audience is physics students who are familiar with the above. I am still amazed that most of relativity is based on geometry, algebra and some postulates which I find remarkable. Comparable to other theories that have changed our views of the universe, the mathematical simplicity of this one I find extremely elegant. Thank you for the comment.
I want you solve one question of relativity I was unable to solve this problem
Book: Robert Resnick Relativity
Page: 48 , Problem:14 , Binary system
Please sir my humble request with you , no one was help me 😥🥺
Sounds interesting! Drop me a comment with the problem as I don't have the book handy at home.
Problem containing figure too so it difficult to share 😅
@@sunnyanandphylo8605 Take a picture and post it on imgur (or any image site) then post the link
@@potato_nugget yeah 👍 thanks wait 👍
@@zhelyo_physics drive.google.com/file/d/1N0xiPx-kuMurYyuQoi9B6ddJ4QJzuOGk/view?usp=drivesdk
This is the problem